Question: The tangent line to the graph of function $g$ at the point $(1,5)$ passes through the point $(3,4)$. Find $g'(1)$. $g'(1)=$
Explanation: The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $g'(1)$ gives the slope of the tangent line to the graph of $g$ where $x=1$, which is the point $(1,5)$. We know this line passes through $(1,5)$, and we are also given that it passes through $(3,4)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{4-5}{3-1} \\\\ &=\dfrac{-1}{2} \end{aligned}$ In conclusion, $g'(1)=-\dfrac{1}{2}$.